Finding the EUI address in an IPv6 address

As this is a homework question, I'm hoping for some guidance in the correct direction or a website. I was given an IPv6 address and I need to find the EUI portion of the address. And EUI is just the first 24 bits of a MAC address with the seventh bit flipped, followed by FFFE (16 bits), and the last 24 bits of the MAC address. Naturally, if you see FFFE in the address anywhere, it's an EUI address.

These are the IPv6 addresses:
2005:007D:4500:000C:0000:0010:11D8:6E71
2005:007D:4500:000C:0000:0002:4A5C:E1C9

The prefix is 2005:7D:4500:C::/64

FFFE isn't in either of them. Did I completly miss something regarding EUI addressing? I did reach out to the professor as it's an online course but I wanted to see if anyone here can identify if something is off with this or if I'm just compleltly missing something. I don't think 0000:0002:4A5C:E1C9 is a valid EUI.

Does not look to be a two way street. That is, it's a method we can use to automatically configure IPv6 host addresses. An IPv6 device will use the MAC address of its interface to generate a unique 64-bit interface ID. However, this doesn't mean that a DHCP assigned address will contain that MAC address.

I think we need to take a step back here. The IP address does not contain the MAC address. However there was a need to increase the MAC address pool for IPv6. "Guidelines for 48-Bit Global Identifier (EUI-48)" (PDF). IEEE Standards Association. IEEE. Retrieved 16 April 2015 gives a great overview of this.

If we assume it's EUI-48, that would make them
0010:11D8:6E71
0002:4A5C:E1C9
Respectivly to the first post

Since it does not contain FFFE or FFFF, it is not an EUI-48 mapped to an EUI-64; therfore, the entire EUI-48 does not need to be the interface ID. I think that's where I was getting stuck on assuming my logic on this is correct. The interface ID's are:
0000:0010:11D8:6E71
0000:0002:4A5C:E1C9